Create Zip File from List<Byte[]> in Memory

Question

I have a web services that returns a list of files. Something like this:

public FileModel(){
  string FileName {get;set;}
  byte[] FileStream {get;set;}
  string FileType {get;set;} 
}

My Service would return:

  List<FileModel> files;

I have to return this list to browser, so I need to compress these files into a zip folder.

However, I cant figure out how to do this, as .NET ZipArchive CreateFromDirectory is requiring me to provide a directory where the file to be zipped are. But I don't have a directory, I just have this list. How can I covert this list to a zipped folder.

What do you mean by zip method? All the constructors of the ZipArchive class accept Stream. — Ivan Stoev
– Ivan Stoev, Commented Apr 20, 2016 at 18:46
I have to return this list to browser, No you can return filename,filetype and filesize to the browser. When user actually requests a specific file, you can return it compressed.... — Eser
– Eser, Commented Apr 20, 2016 at 18:51
@Eser, No, I actually need to return a zipped folder of the files to the browser/user. — Mark
– Mark, Commented Apr 20, 2016 at 19:07
Iconic's DotNetZip (Freeware) allows you to pass a byte[] to create a zip file. I don't know of anything that will let you pass a List<byte()> though. — Kevin
– Kevin, Commented Apr 20, 2016 at 19:11
It also lets you pass a stream or byte[] to the method for creating a file within the zip file. See if your library's CreateEntry method has an override that accepts a stream, if so create a MemoryStream from the byte[] and pass in the stream. — Kevin
– Kevin, Commented Apr 20, 2016 at 19:13

Nkosi · Accepted Answer · 2018-05-24 18:40:41Z

4

Given

public FileModel(){
    string FileName {get;set;}
    byte[] FileStream {get;set;}
    string FileType {get;set;} 
}

The following was written to create a zip file

static class FileModelCompression {

    public static Stream Compress(this IEnumerable<FileModel> files) {
        if (files.Any()) {
            var ms = new MemoryStream();
            using(var archive = new ZipArchive(ms, ZipArchiveMode.Create, leaveOpen: true)) {
                foreach (var file in files) {
                    var entry = archive.add(file);
                }
            }// disposal of archive will force data to be written to memory stream.
            ms.Position = 0; //reset memory stream position.
            return ms;
        }
        return null;
    }

    private static ZipArchiveEntry add(this ZipArchive archive, FileModel file) {
        var entry = archive.CreateEntry(file.FileName, CompressionLevel.Fastest);
        using (var stream = entry.Open()) {
            file.FileStream.CopyTo(stream);
        }
        return entry;
    }
}

edited May 24, 2018 at 18:40

answered Apr 21, 2016 at 20:25

Nkosi

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1 Comment

ProblemAnswerQue Over a year ago

If we use your code, we are able to create this file and see the files in the zip but: When we try to extract the zip, we have an error complaining about an unexpected end of archive.

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